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Ipatiy [6.2K]
4 years ago
14

3. A 2,000-kilogram railroad car moving at 5 m/sec to the east collides with a 6,000-kilogram railroad car moving at 3 m/sec to

the west. If the cars couple together, what is their velocity after the collision?
Physics
1 answer:
Nat2105 [25]4 years ago
5 0
<span>So we wan't to know what is the velocity after a collision of two railroad cars, one moving to the east and the other moving to the west if m1=2000kg, v1=5m/s and m2=6000kg, v2=3m/s. We can find the solution using the law of conservation of momentum for plastic collisions that states that the momentum must remain constant before (left side of the equation) and after (right side of the equation) the collision: m1*v1+m2*v2=(m1+m2)*v. So now we simply plug in the numbers and get: 2000kg * 5m/s + 6000kg * 3m/s = (2000kg + 6000kg)*v. Now we can write: 10000 kgm/s + 18000 kgm/s = 8000kg * v. To get v, the velocity of both railroad cars after the collision we simply divide both sides of the equation with 8000 kg: so v=3.5m/s to the west. </span>
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What is the force needed to give a .25 kg arrow an acceleration of 196m/s2?
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<span>49N is the force needed to give a .25 kg arrow an acceleration of 196m/s2. F =ma ⇒ =( 0.25kg)(196m/s2) = 49N if the arrow is shot horizontally where the applied force is entirely in the x-direction.</span>
5 0
3 years ago
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Lake Preston's latitude is about 44.4 degrees north of the equator. Which global wind belt are
Serggg [28]

Answer:

Idk

Explanation:

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5 0
3 years ago
A boat is drifting with a speed of 5.0\,\dfrac{\text m}{\text s}5.0 s m ​ 5, point, 0, space, start fraction, m, divided by, s,
zepelin [54]

Answer:

d=42m(-\hat{i})

Explanation:

For this problem, we need to apply the formulas of constant accelerated motion.

To obtain the boat displacement we need to calculate the displacement because of the river flow and the displacement done because of the boat motor.

for the river:

d_r=v*t\\d_r=5m/s*6s\\d_r=30m(\hat{i})

for the boat:

x=\frac{1}{2}*a*t^2\\\\x=\frac{1}{2}*4.0m/s*(6s)^2\\\\\\x=72m(-\hat{i})

So the final displacement is given by:

d=dr+x\\d=30m-72m\\d=42m(-\hat{i})

8 0
4 years ago
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: The maximum theoretical efficiency of a Carnot engine operating between reservoirs at the steam point and at room temperature
Hunter-Best [27]

Answer:

The value is   \eta  =  0.2145  or  21.45%

Explanation:

From the question we are told that

    The first reservoir is at steam point  T_s =  100^o C =  100 + 273 = 373 \ K  

    The  second reservoir is at room temperature T_r  =  20^o C = 293 \ K

Generally the  maximum theoretical efficiency of a Carnot engine  is mathematically evaluated as

     \eta  =  1- \frac{T_r}{T_s}

=>    1 - \frac{ 293}{373}

=>    \eta  =  0.2145

5 0
3 years ago
An object of mass 8kg moved around the circle of radius 4m. with a constant speed of 15m/s.
Marianna [84]

Answer:

a. Angular velocity = 0.267rad/s.

b. Centripetal acceleration = 56.25m/s.

Explanation:

<u>Given the following data;</u>

Mass, m = 8kg

Radius, r = 4m

Constant speed, V = 15m/s

a. To find the angular velocity

Angular velocity = radius/speed

Substituting into the equation, we have;

Angular velocity = 4/15

Angular velocity = 0.267rad/s

b. To find the acceleration;

Centripetal acceleration = V²/r

Substituting into the equation, we have;

Centripetal acceleration = 15²/4

Centripetal acceleration = 225/4

Centripetal acceleration = 56.25m/s.

4 0
3 years ago
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